A bullet of mass m = 50 gm strikes a sand bag of mass M = 5 kg hanging from a fixed point, with a horizontal velocity `vecv_(p)` . If bullet sticks to the sand bag then the ratio of final & initial kinetic energy of the bullet is (approximately) :

Explanation

Since the bullet strikes the sand bag with a horizontal velocity of magnitude `v_(p)`. Therefore, its momentum just before the impact `= p_(1)=mV_(p)`.
Let the speed of the combination (sand bag + bullet) just after the impact be v.. Therefore the momentum of the system (sand bag + bullet) just after the impact `= p_(f)=(M+m)v.`. Since no net horizontal external force acts on the system (M+m) during the collision, its net momentum remains conserved during the impact
`rArr p_(i)=p_(f)`
`rArr mv=(M+m)v.`
`rArr v.=((m)/(M+m))v`. Since `m lt lt M, v.=(mv)/(M) " "` ...(1)
Fraction of KE lost by the bullet during impact
`eta=((1)/(2)mv.^(2))/((1)/(2)mv^(2))=((v.)/(v))^(2) " "` ....(2)
Using (1) and (2), we obtain,
`eta =((m)/(M))^(2)=((1//20)/(5))^(2)=10^(-4)`